Documents

POTWE-13-AE-23-S

Description
Elementary Mathematics
Categories
Published
of 2
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the WeekProblem E and SolutionThese are Sum Boxes Problem In a sequence of 12 numbers, each number after the first three is equal to thesum of the previous three numbers. The 3 rd number in the sequence is 6, the6 th number in the sequence is 11, and the 11 th number in the sequence is 14.Determine all 12 numbers in the sequence. Solution Let  a 1  be the first number in the sequence,  a 2  be the second,  a 4  be the fourth,and so on, until  a 12  which is the 12 th number in the sequence. The 12 boxes arelabelled in the following diagram. 6 11 14a 1  a 2  a 4  a 5  a 7  a 8  a 9  a 10  a 12 Each number after the third number is equal to the sum of the previous threenumbers. Therefore, looking at the 6 th term, we have 11 = 6 +  a 4  +  a 5  or a 4  +  a 5  = 5.Looking at the 7 th term,  a 7  =  a 4  +  a 5  + 11 = 5 + 11 = 16, since  a 4  +  a 5  = 5. 6 11 14a 1  a 2  a 4  a 5  16 a 8  a 9  a 10  a 12 Looking at the 9 th term,  a 9  = 11 + 16 +  a 8  = 27 +  a 8 .Looking at the 10 th term,  a 10  = 16 +  a 8  +  a 9  = 16 + ( a 8 ) +( a 8  +27) = 2 a 8  +43.Looking at the 11 th term, a 11  =  a 8  +  a 9  +  a 10  = ( a 8 ) + ( a 8  + 27) + (2 a 8  + 43) = 4 a 8  + 70.We are given that the 11 th term is  a 11  = 14.Therefore, 4 a 8  + 70 = 14, or 4 a 8  = − 56, or  a 8  = − 14.Therefore,  a 9  =  a 8  + 27 = − 14 + 27 = 13. a 10  = 2 a 8  + 43 = 2( − 14) + 43 = 15.Also,  a 12  =  a 9  +  a 10  +  a 11  = 13 + 15 + 14 = 42.  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING So far, we know the sequence is 6 11 14a 1  a 2  a 4  a 5  16 -14 13 15 42 Working backwards,  − 14 =  a 5  + 11 + 16, so  a 5  = − 41.From earlier,  a 4  +  a 5  = 5 and  a 5  = − 41, so  a 4  = 46.Continuing backwards,  a 5  =  a 2  + 6 +  a 4 , so  − 41 =  a 2  + 6 + 46, or  a 2  = − 93.And finally,  a 4  =  a 1  +  a 2  + 6, so 46 =  a 1  + ( − 93) + 6, or  a 1  = 133.Therefore, the sequence of 12 numbers is133 ,  − 93 ,  6 ,  46 ,  − 41 ,  11 ,  16 ,  − 14 ,  13 ,  15 ,  14 ,  42 6 11 14133-93 46 16 -14 13 15 42-41 We can indeed check that in this sequence, each number after the first threenumbers is equal to the sum of the previous three numbers.

Ross 7e Ch16

Jul 23, 2017

Ethics

Jul 23, 2017
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks